Question

Two cyclists at the same time went from A and B (one from each point) points, they drive with constant velocity. When they arrive at any point, they immedeatelly turn around and drive back. First time they had met 40km from B, and second time, after 8h from the first meet up, they met 20km from point A. Find the distance between A and B, velocity of the cyclists.

Answer 1

1 meet ->
2 meet -> OR OR

Those three possibilities of directions bug me the most. The problem is that the cyclist could have driven 3 different distances until they met. Is that really the right way to look at the problem?

Answer 2

Sorry, actually there can be 4 different possibilities of directions (hence, the distances).

@ganeshie8, maybe you have any idea where to start with this one?

Answer 3

@Kainui, maybe you have any ideas?

Answer 4

Let the time when the cycles first meet be t. then
distance traveled by cycle A be \(v_at\) & distance traveled by cycle B be is 40 km.

the time of second meet is (t+8).
Net distance traveled by cycle B after time (t+8) is \(40+v_at+20\) km.
net distance traveled by cycle A after time (t+8) is \(v_a(t+8)\) km

Answer 5

@jiteshmeghwal9 couldn't it be possible that B traveled 40 + v_a * t, if they were slow? So, it would be that at 20km from A cyclist A caught up to them.

Answer 6

Is answer approximately 111 km ?

Answer 7

@Zyberg

Answer 8

@jiteshmeghwal9 I don't have answers to the problems, but I reckon that it should be a whole integer, without any approximations. Might be wrong, but so far every answer has been an integer.